### 3.225 $$\int \frac{e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^6} \, dx$$

Optimal. Leaf size=125 $-\frac{\left (a+\frac{1}{x}\right )^3}{7 a^4 c^6 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}}+\frac{24 \left (a+\frac{1}{x}\right )^2}{35 a^3 c^6 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}-\frac{46 \left (a+\frac{1}{x}\right )}{35 a^2 c^6 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}+\frac{35 a+\frac{13}{x}}{35 a^2 c^6 \sqrt{1-\frac{1}{a^2 x^2}}}$

[Out]

(-46*(a + x^(-1)))/(35*a^2*c^6*(1 - 1/(a^2*x^2))^(3/2)) + (24*(a + x^(-1))^2)/(35*a^3*c^6*(1 - 1/(a^2*x^2))^(5
/2)) - (a + x^(-1))^3/(7*a^4*c^6*(1 - 1/(a^2*x^2))^(7/2)) + (35*a + 13/x)/(35*a^2*c^6*Sqrt[1 - 1/(a^2*x^2)])

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Rubi [A]  time = 0.410555, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.278, Rules used = {6175, 6178, 852, 1635, 637} $-\frac{\left (a+\frac{1}{x}\right )^3}{7 a^4 c^6 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}}+\frac{24 \left (a+\frac{1}{x}\right )^2}{35 a^3 c^6 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}-\frac{46 \left (a+\frac{1}{x}\right )}{35 a^2 c^6 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}+\frac{35 a+\frac{13}{x}}{35 a^2 c^6 \sqrt{1-\frac{1}{a^2 x^2}}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/(E^(3*ArcCoth[a*x])*(c - a*c*x)^6),x]

[Out]

(-46*(a + x^(-1)))/(35*a^2*c^6*(1 - 1/(a^2*x^2))^(3/2)) + (24*(a + x^(-1))^2)/(35*a^3*c^6*(1 - 1/(a^2*x^2))^(5
/2)) - (a + x^(-1))^3/(7*a^4*c^6*(1 - 1/(a^2*x^2))^(7/2)) + (35*a + 13/x)/(35*a^2*c^6*Sqrt[1 - 1/(a^2*x^2)])

Rule 6175

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[d^p, Int[u*x^p*(1 + c/(d*
x))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c^2 - d^2, 0] &&  !IntegerQ[n/2] && Inte
gerQ[p]

Rule 6178

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^n, Subst[Int[((c
+ d*x)^(p - n)*(1 - x^2/a^2)^(n/2))/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

Rule 852

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[((f + g*x)^n*(a + c*x^2)^(m + p))/(d - e*x)^m, x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 637

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-(a*e) + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rubi steps

\begin{align*} \int \frac{e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^6} \, dx &=\frac{\int \frac{e^{-3 \coth ^{-1}(a x)}}{\left (1-\frac{1}{a x}\right )^6 x^6} \, dx}{a^6 c^6}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1-\frac{x}{a}\right )^3 \left (1-\frac{x^2}{a^2}\right )^{3/2}} \, dx,x,\frac{1}{x}\right )}{a^6 c^6}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (1+\frac{x}{a}\right )^3}{\left (1-\frac{x^2}{a^2}\right )^{9/2}} \, dx,x,\frac{1}{x}\right )}{a^6 c^6}\\ &=-\frac{\left (a+\frac{1}{x}\right )^3}{7 a^4 c^6 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}}+\frac{\operatorname{Subst}\left (\int \frac{\left (1+\frac{x}{a}\right )^2 \left (3 a^4+7 a^3 x+7 a^2 x^2+7 a x^3\right )}{\left (1-\frac{x^2}{a^2}\right )^{7/2}} \, dx,x,\frac{1}{x}\right )}{7 a^6 c^6}\\ &=\frac{24 \left (a+\frac{1}{x}\right )^2}{35 a^3 c^6 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}-\frac{\left (a+\frac{1}{x}\right )^3}{7 a^4 c^6 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}}-\frac{\operatorname{Subst}\left (\int \frac{\left (1+\frac{x}{a}\right ) \left (33 a^4+70 a^3 x+35 a^2 x^2\right )}{\left (1-\frac{x^2}{a^2}\right )^{5/2}} \, dx,x,\frac{1}{x}\right )}{35 a^6 c^6}\\ &=-\frac{46 \left (a+\frac{1}{x}\right )}{35 a^2 c^6 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}+\frac{24 \left (a+\frac{1}{x}\right )^2}{35 a^3 c^6 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}-\frac{\left (a+\frac{1}{x}\right )^3}{7 a^4 c^6 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}}+\frac{\operatorname{Subst}\left (\int \frac{39 a^4+105 a^3 x}{\left (1-\frac{x^2}{a^2}\right )^{3/2}} \, dx,x,\frac{1}{x}\right )}{105 a^6 c^6}\\ &=-\frac{46 \left (a+\frac{1}{x}\right )}{35 a^2 c^6 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}+\frac{24 \left (a+\frac{1}{x}\right )^2}{35 a^3 c^6 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}-\frac{\left (a+\frac{1}{x}\right )^3}{7 a^4 c^6 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}}+\frac{35 a+\frac{13}{x}}{35 a^2 c^6 \sqrt{1-\frac{1}{a^2 x^2}}}\\ \end{align*}

Mathematica [A]  time = 0.0705729, size = 66, normalized size = 0.53 $\frac{x \sqrt{1-\frac{1}{a^2 x^2}} \left (8 a^4 x^4-24 a^3 x^3+20 a^2 x^2+4 a x-13\right )}{35 c^6 (a x-1)^4 (a x+1)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/(E^(3*ArcCoth[a*x])*(c - a*c*x)^6),x]

[Out]

(Sqrt[1 - 1/(a^2*x^2)]*x*(-13 + 4*a*x + 20*a^2*x^2 - 24*a^3*x^3 + 8*a^4*x^4))/(35*c^6*(-1 + a*x)^4*(1 + a*x))

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Maple [A]  time = 0.043, size = 66, normalized size = 0.5 \begin{align*}{\frac{ \left ( 8\,{x}^{4}{a}^{4}-24\,{x}^{3}{a}^{3}+20\,{a}^{2}{x}^{2}+4\,ax-13 \right ) \left ( ax+1 \right ) }{35\, \left ( ax-1 \right ) ^{5}{c}^{6}a} \left ({\frac{ax-1}{ax+1}} \right ) ^{{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^6,x)

[Out]

1/35*((a*x-1)/(a*x+1))^(3/2)*(8*a^4*x^4-24*a^3*x^3+20*a^2*x^2+4*a*x-13)*(a*x+1)/(a*x-1)^5/c^6/a

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Maxima [A]  time = 1.05046, size = 131, normalized size = 1.05 \begin{align*} \frac{1}{560} \, a{\left (\frac{35 \, \sqrt{\frac{a x - 1}{a x + 1}}}{a^{2} c^{6}} + \frac{\frac{28 \,{\left (a x - 1\right )}}{a x + 1} - \frac{70 \,{\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac{140 \,{\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} - 5}{a^{2} c^{6} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{7}{2}}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^6,x, algorithm="maxima")

[Out]

1/560*a*(35*sqrt((a*x - 1)/(a*x + 1))/(a^2*c^6) + (28*(a*x - 1)/(a*x + 1) - 70*(a*x - 1)^2/(a*x + 1)^2 + 140*(
a*x - 1)^3/(a*x + 1)^3 - 5)/(a^2*c^6*((a*x - 1)/(a*x + 1))^(7/2)))

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Fricas [A]  time = 1.5806, size = 204, normalized size = 1.63 \begin{align*} \frac{{\left (8 \, a^{4} x^{4} - 24 \, a^{3} x^{3} + 20 \, a^{2} x^{2} + 4 \, a x - 13\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{35 \,{\left (a^{5} c^{6} x^{4} - 4 \, a^{4} c^{6} x^{3} + 6 \, a^{3} c^{6} x^{2} - 4 \, a^{2} c^{6} x + a c^{6}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^6,x, algorithm="fricas")

[Out]

1/35*(8*a^4*x^4 - 24*a^3*x^3 + 20*a^2*x^2 + 4*a*x - 13)*sqrt((a*x - 1)/(a*x + 1))/(a^5*c^6*x^4 - 4*a^4*c^6*x^3
+ 6*a^3*c^6*x^2 - 4*a^2*c^6*x + a*c^6)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))**(3/2)/(-a*c*x+c)**6,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}}{{\left (a c x - c\right )}^{6}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^6,x, algorithm="giac")

[Out]

integrate(((a*x - 1)/(a*x + 1))^(3/2)/(a*c*x - c)^6, x)